A body weighs 64N on the surface of the earth. what is the gravitational force (in N) on it due to the earth at a height equal to one-third of the radius of the earth?

Here, mg = 64 N; 

At h=R/3

The value of acceleration due to gravity at a height h( when h is not negligible as compared to R),
$\begin{array}{l}{g}^{\text{'}}=g\frac{R^2}{{(R+h)}^2}\\ orm{g}^{\text{'}}=mg\frac{R^2}{{(R+h)}^2}\end{array}$

$=64\times \frac{R^2}{{(R+R/3)}^2}=\frac{9}{16}\times 64=36N$